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Is it a valid inference (in, say, any quantified modal logic S5 system) to go from ◊(∃x)◻Fx and ◻(∀x)(Fx --> E!x) to ◊(∃x)◻(Fx & E!x)? (E!x is the existence predicate, by the way.)

  • If "E!x" is the existence predicate what is "∃x"? Does it refer to "unique" existence? Also is there a text where this problem is presented? This would provide context. Welcome to Philosophy! – Frank Hubeny Dec 29 '18 at 2:35
  • The question does not make any sense. S5, QML, and SQML are totally different systems; therefore, it does not make sense to say any QML S5 system. That said, in S5 the statement is obviously not derivable since the syntax does not include quantification. Also, QML and SQML do not have unique existence in their syntax; therefore, it won't be derivable in those systems either. – Bertrand Wittgenstein's Ghost Dec 31 '18 at 6:10
  • @BertrandWittgenstein'sGhost QML+S5 does make sense. S5 is generally thought of as a characterisation of the accessibility relation, not so much as an axiomatic system in propositional logic. The symbolism of E!x does not mean unique quantification (there is no scope), but rather existence. In some contexts it is helpful to use a common domain for all worlds, since it is otherwise not so clear how to interpret quantification into modal contexts, and if one wants to allow that things "possibly don't exist" one can introduce an existence predicate to signify actualisation in that world. – Jishin Noben Jan 13 at 19:22
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I am assuming, that there is a domain of objects that is common to all worlds, and the existence predicate E! signifies actualisation in each world (although that does not matter).

Your inference is valid. In S5 accessibility is an equivalence relation. From your first premise you get an object a which in all (here S5 was used) worlds lies in the extension of F. By your second premise, it also E!-exists in all worlds. Therefore (∃x)◻(Fx & E!x) and since we are in S5, ◊(∃x)◻(Fx & E!x).

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